The length of a rectangle is 2cm greater than the width. The area is 80cm2. Find the length and width.
L=W+2
W=W
[tex]\begin{gathered} A=L\cdot W \\ A=(W+2)\cdot W \\ A=W^2+2W \\ A=80\operatorname{cm} \\ Then, \\ 80=W^2+2W \\ W^2+2W-80=0 \end{gathered}[/tex][tex]\Delta=4+320=324[/tex][tex]\begin{gathered} W=\frac{-2\pm\sqrt[]{324}}{2}=\frac{-2\pm18}{2} \\ W_1=\frac{-20}{2}=-10 \\ W_2=\frac{16}{2}=8 \end{gathered}[/tex]The width should be positive, therefore W=8
L=W+2
L=8+2=10
The length is L=10
